"Completion property" in $(\beta\omega,+)$

Question

Let $\beta\omega$ be collection of all ultrafilters on $\omega$ (principal and non-principal). We endow $\beta\omega$ with an operation $+$ in the following way. For ${\bf a}, {\bf b}\in \beta\omega$, set $${\bf a}+{\bf b} = \big\{ N\subseteq \omega:\{x \in \omega:\{y\in\omega: x+y\in N\}\in {\bf b}\}\in {\bf a}\big\}.$$

Standard exercises show that ${\bf a}+{\bf b}\in \beta\omega$, and that the operation $+:\beta\omega \times \beta\omega \to \beta\omega$ is associative.

If we identify a member $n\in\omega$ with the principal ultrafilter containing the singleton $\{n\}$ as an element, the addition just introduced agrees with the addition given in $\omega$. In $\omega$, we have the following "completion property": given $a, b\in \omega$, there is $x\in \omega$ such that $a+x=b$ or $b+x=a$.

This motivates the following question: if ${\bf a},{\bf b} \in \beta\omega$, is there ${\bf x}\in\beta\omega$ with ${\bf a}+{\bf x} = {\bf b}$ or ${\bf b}+{\bf x} = {\bf a}$?

Answer 1

No. Kunen proved that there are many weak P-points in $\beta\omega-\omega$, which means they are nonprincipal ultrafilters that are not in the closure of any countable set of other nonprincipal ultrafilters. In particular, a weak P-point is never a sum of nonprincipal ultrafilters because $\mathbf a+\mathbf b$ is in the closure of the set $\{\mathbf b+n:n\in\omega\}$.

Answer 2

Write $(\alpha x)\ P(x)$ for the statement $\{x:P(x)\}\in\alpha$. Now suppose that the set $\{1,2,4,8,\dots\}\in\alpha$. If $\beta+\gamma=\alpha$, then $(\beta x)\ (\gamma y)\ x+y\in\{1,2,4,8,\dots\}$. In particular, if $\beta$ is non-principal, then there must exist $x_1\ne x_2$ such that $(\gamma y)\ x_1+y\in\{1,2,4,8,\dots\}$ and $(\gamma y)\ x_2+y\in\{1,2,4,8,\dots\}$. But the set of $y$ with that property is finite, so this can't happen unless $\gamma$ is principal.

To put it a different way, if $\alpha$ is a sum $\beta+\gamma$ of two non-principal ultrafilters and $A\in\alpha$, then many shifts of $A$ must belong to $\gamma$, so their finite intersections must belong to $\gamma$. It's easy to find infinite sets $A$ for which this doesn't happen, and it's also easy to make sure they belong to $\alpha$.